Let the point L lying in the first quadrant b | vedclass

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(A) Given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$. Here,$a^2=4, b^2=3$,so $a=2, b=\sqrt{3}$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sq

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$\frac{\sqrt{5}}{4}$

Vedclass · Answer

(A) Given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$. Here,$a^2=4, b^2=3$,so $a=2, b=\sqrt{3}$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$. Since $L$ lies in the first quadrant,the coordinates of $L$ are $(ae, \frac{b^2}{a}) = (2 	imes \frac{1}{2}, \frac{3}{2}) = (1, \frac{3}{2})$. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $x_1 = 1, y_1 = \frac{3}{2}, a^2 = 4, b^2 = 3$: $\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3 \Rightarrow 4x - 2y = 1$. The normal meets the major axis ($x$-axis) at $P$ by setting $y=0$: $4x = 1 \Rightarrow x = \frac{1}{4}$. So,$P = (\frac{1}{4}, 0)$. The normal meets the minor axis ($y$-axis) at $Q$ by setting $x=0$: $-2y = 1 \Rightarrow y = -\frac{1}{2}$. So,$Q = (0, -\frac{1}{2})$. The distance $PQ = \sqrt{(\frac{1}{4} - 0)^2 + (0 - (-\frac{1}{2}))^2} = \sqrt{\frac{1}{16} + \frac{1}{4}} = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$.

Let the point $L$ lying in the first quadrant be one end of a latus rectum of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $P$ and $Q$ be the points where the normal drawn at $L$ to this given ellipse meets the major axis and the minor axis. Then the distance between $P$ and $Q$ is

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